Let $(X_t)$ be a real continuous stochastic process such that $E[ \int_0^1 (\frac{X_t}{1-t})^2 dt ] <\infty$. Let $f_t$ be the probability density function (PDF) of $(X_t)$.
Since $Y_t= \frac{X_t}{1-t}$ is $L^2$ in time and space, and that the measure of $\Omega \times [0,1]$ is finite, $(Y_t)$ is also $L^1$ in time and space, and hence,
$$I:=E\left[ \int_0^1 \frac{X_t}{1-t} dt \right] <\infty.$$
I would like to know if the following integral is equal to $I$:
$$\int_\mathbb R x \left(\int_0^1 \frac{f_t(x)}{1-t} dt\right) d x,$$
and if so, is $\int_0^1 \frac{f_t(x)}{1-t} dt$ the PDF of the random variable $\int_0^1 \frac{X_t}{1-t} dt$ ? Is the function $x \rightarrow x\int_0^1 \frac{f_t(x)}{1-t} dt$ guaranteed to be integrable in $x$ ?
I obtained this integral by following these steps:
$$I = \int_0^1 \frac{E[X_t]}{1-t} dt = \int_0^1 \frac{ \int_\mathbb R x f_t(x) dx}{1-t} dt =\int_\mathbb R x \left(\int_0^1 \frac{f_t(x)}{1-t} dt\right) d x $$
I used Fubini two times, and I am not sure if knowing $I < \infty$ is enough to use Fubini in these ways.
Best Answer
Your first application of Fubini's theorem is correct, but the second one does not seem to hold in general.
First application
Recall that to apply Fubini's theorem to $f$, you must have $f \in L^1( \Omega \times [0,1], \mathbb{P} \times \lambda)$. Under your assumptions, this holds, since $$\int_{\Omega \times [0,1]} \left| \frac{X_t}{1-t} \right| d (\mathbb{P} \times \lambda) \leq \int_{\Omega \times [0,1]} \left| \frac{X_t}{1-t} \right|^2 d (\mathbb{P} \times \lambda) = \mathbb{E} \int_0^1 \frac{X_t^2}{(1-t)^2}dt < \infty $$ where the inequality is Hölder's inequality and the equality is Tonelli's theorem (which allows you to exchange iterated integrals for positive functions). It follows that $$\mathbb{E} \int_0^1 \frac{X_t}{1-t} dt = \int_0^1 \frac{\mathbb{E} (X_t)}{1-t} = \int_0^1 \frac{\int_{\mathbb{R} } x f_t(x) }{1-t}$$ by Fubini's theorem.
Second application
Your second application doesn't seem to hold. For a counterexample, consider the time-changed and time-reversed Brownian motion $X_t = B_{(1-t)^2}$, whose density is given by $$f_t(x) = \frac{1}{\sqrt{2 \pi} (1-t) } \exp \left( - \frac{x^2}{2(1-t)^2} \right)$$ Observe that your moment condition holds: $$\mathbb{E} \int_0^1 \frac{X_t^2}{(1-t)^2}dt = \int_0^1 \frac{\mathbb{E}X_t^2}{(1-t)^2}dt = \int_0^1 \frac{(1-t)^2}{(1-t)^2} = 1 < \infty$$ So Fubini's theorem tells us that: $$\mathbb{E} \int_0^1 \frac{X_t}{1-t} dt = \int_0^1 \frac{\mathbb{E} X_t}{1-t} dt= 0$$ However, according to WolframAlpha, we get: $$\int_0^1 \frac{f_t(x)}{1-t} = \sqrt{ \frac{\pi}{2} } \left( \frac{1}{x} - \mathrm{erf} \left( \frac{x}{\sqrt{2}} \right) \right)$$ whence, $$\int_{\mathbb{R}} x \left(\sqrt{ \frac{\pi}{2} } \left( \frac{1}{x} - \mathrm{erf} \left( \frac{x}{\sqrt{2}} \right) \right) \right)dx $$ does not exist.